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The electric field in a region is given by overrightarrow E=(2 i+4 j+6 k) × 10^3 ~N / C. The flux of the field through a rectangular surface parallel…

JEE Main 2025 — Physics Electromagnetism

2025integereasy

The electric field in a region is given by E→=(2i^+4j^+6k^)×103 N/C\overrightarrow{\mathrm{E}}=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times 10^3 \mathrm{~N} / \mathrm{C}E=(2i^+4j^​+6k^)×103 N/C. The flux of the field through a rectangular surface parallel to x−zx-zx−z plane is 6.0Nm2C−16.0 \mathrm{Nm}^2 \mathrm{C}^{-1}6.0Nm2C−1. The area of the surface is ________ cm2\mathrm{cm}^2cm2.

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Held on 7 Apr 2025 · Verified 6 Jul 2026.

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